Evaluate $~~\int \sin^{-1}(x)\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $x\sin^{-1}(x)+\sqrt{1-x^2}+C$ (Choice B) B $\sin^{-1}(x)+\sqrt{1-x^2}+C$ (Choice C) C $\dfrac{x}2\sin^{-1}(x)+\sqrt{1-x^2}+C$ (Choice D) D $x\sin^{-1}(x)+\dfrac{x}{\sqrt{1-x^2}}+C$
We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \sin^{-1}x$ and $~dv=dx\,$. Then $~du = \dfrac{1}{\sqrt{1-x^2}}\, dx~$ and $~v = \int dx = x\,$. Integration by parts gives $ \int \sin^{-1}x\,dx = x\sin^{-1}x-\int\dfrac{x}{\sqrt{1-x^2}}\,dx$ We can determine this second integral using a $~u$ -substitution. Let $~u=1-x^2$. Then $~du=-2x\,dx\,$, or $~-\dfrac12du=x\, dx\,$. Now we have $~~\int\dfrac{x}{\sqrt{1-x^2}}\,dx=\int-\dfrac12\cdot\dfrac1{\sqrt{u}}\, du=-\sqrt{u}+C\,$. Our final result is now $ \int \sin^{-1}x\,dx = x\sin^{-1}x-(-\sqrt{u}+C) =x\sin^{-1}x+\sqrt{1-x^2}+C\,$.